Question

Assertion (A): $$ \frac{d}{dx} \left( \frac{x^2 \sin x}{\log x} \right) = \frac{x^2 \cos x \cdot \log x - x^2 \sin x \cdot \frac{1}{x}}{(\log x)^2} + \frac{2x \sin x}{\log x} \Rightarrow \text{(matches the structure of product + quotient rule)} $$ Reason (R): $$ \frac{d}{dx} \left( \frac{uv}{w} \right) = \frac{uv'}{w} + \frac{u'v}{w} - \frac{uvw'}{w^2} \Rightarrow \text{(correct general derivative formula)} $$

• A. A is true, R is true and R is correct explanation of A
• B. A is true, R is true but not correct explanation
• C. A is true, R is false
• D. A is false, R is true

Hint:

When differentiating a fraction of products, use the formula: \[ \frac{d}{dx} \left( \frac{uv}{w} \right) = \frac{uv'}{w} + \frac{u'v}{w} - \frac{uvw'}{w^2} \]

Answer 1

The Correct Option is A

We are differentiating: \[ f(x) = \frac{x^2 \sin x}{\log x} \Rightarrow \text{This is } \frac{uv}{w},\ u = x^2,\ v = \sin x,\ w = \log x \] Apply the derivative rule as given in reason (R): \[ \frac{d}{dx} \left( \frac{uv}{w} \right) = \frac{uv'}{w} + \frac{u'v}{w} - \frac{uvw'}{w^2} \Rightarrow \frac{x^2 \cos x + 2x \sin x}{\log x} - \frac{x^2 \sin x \cdot \frac{1}{x}}{(\log x)^2} = \frac{x^2 \cos x + 2x \sin x}{\log x} - \frac{x \sin x}{(\log x)^2} \] Which simplifies and matches assertion. Hence both A and R are true and R is the correct explanation of A.